Question

If an object is moving at `12 m/s` over a surface with a kinetic friction coefficient of `u_k=5 /g`, how far will the object continue to move?

Answer 1

The distance is `=14.4m`

Explanation:

Let the mass of the object be `=mkg`

`v=12ms^-1`

The coefficient of kinetic friction is

`mu_k=F_r/N`

`N=mg`

So,

`F_r=mu_k*N`

`=5/g*mg=5m N`

The work done by `F_r` is equal to the kinetic energy of the object.

`F_r*d=1/2mv^2`

`d=(1/2mv^2)/(5m)=v^2/10=12^2/10=144/10=14.4m`